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Heston Model: Calibration and Simulation
Hard
·
28 min read
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Interactive lab
Derivatives Pricing
Stochastic Volatility
Heston Model
Fourier Pricing
Calibration
1
Article
2
Lab
3
Quiz
Quick Quiz
1.
For
d
v
t
=
κ
(
θ
−
v
t
)
d
t
+
ξ
v
t
d
W
t
(
2
)
dv_t=\kappa(\theta-v_t)\,dt+\xi\sqrt{v_t}\,dW_t^{(2)}
d
v
t
=
κ
(
θ
−
v
t
)
d
t
+
ξ
v
t
d
W
t
(
2
)
, the Feller condition is
2
κ
θ
>
ξ
2
2\kappa\theta>\xi^2
2
κ
θ
>
ξ
2
. What does its violation imply?
The log-price characteristic function ceases to exist
The variance can reach zero with positive probability
The correlation
ρ
\rho
ρ
must be set to zero
The SDE is ill-posed and has no solution
2.
In
φ
T
(
u
)
=
exp
(
i
u
r
T
+
A
(
u
,
T
)
+
B
(
u
,
T
)
v
0
)
\varphi_T(u)=\exp(iurT+A(u,T)+B(u,T)v_0)
φ
T
(
u
)
=
exp
(
i
u
r
T
+
A
(
u
,
T
)
+
B
(
u
,
T
)
v
0
)
, why do
A
A
A
and
B
B
B
satisfy ODEs?
Complete: every contingent claim can be replicated
The log-price is exactly Gaussian at all horizons
The state process
(
S
t
,
v
t
)
(S_t,v_t)
(
S
t
,
v
t
)
is a Markov diffusion
Affine: the generator on
e
A
+
B
v
e^{A+Bv}
e
A
+
B
v
gives
v
v
v
-linear terms
3.
The full-truncation Euler scheme uses
v
t
+
=
max
(
v
t
,
0
)
v_t^+=\max(v_t,0)
v
t
+
=
max
(
v
t
,
0
)
inside the drift/diffusion. What is its key advantage over the 'absorption' scheme (which clamps
v
t
+
Δ
t
≥
0
v_{t+\Delta t}\ge 0
v
t
+
Δ
t
≥
0
after each step)?
It yields a smaller variance for the Monte Carlo estimator
It makes the scheme satisfy the Feller condition exactly
It lets variance go negative and recover, reducing upward bias
It is faster because it avoids taking a maximum each step
4.
In Levenberg-Marquardt calibration of Heston,
κ
\kappa
κ
and
θ
\theta
θ
are individually well-identified from a standard implied-vol surface, even when
κ
θ
\kappa\theta
κ
θ
is held fixed.
True
False
5.
Why is the Albrecher et al. (2007) formulation of the Heston CF preferred over the original Heston (1993) form for long maturities?
It converges faster in the Fourier pricing integral
It removes the need for the Feller condition
The original lands on the wrong Riemann sheet (branch cuts)
The original CF is only valid for
T
<
1
T<1
T
<
1
year
6.
In the Lewis (2001) formula for the Heston call, which integration contour is used, and why does it avoid a dampening parameter
α
\alpha
α
?
The contour
I
m
(
u
)
=
+
1
\mathrm{Im}(u)=+1
Im
(
u
)
=
+
1
; it avoids the pole at
u
=
0
u=0
u
=
0
The real axis; no dampening is needed because the Heston CF decays rapidly
The imaginary axis; the CF is real-valued there
The contour
I
m
(
u
)
=
−
1
/
2
\mathrm{Im}(u)=-1/2
Im
(
u
)
=
−
1/2
; the payoff has a simple transform along this strip
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